Problem: $\overline{AB}$ = $10$ $\overline{BC} = {?}$ $A$ $C$ $B$ $10$ $?$ $ \sin( \angle BAC ) = \frac{4}{5}, \cos( \angle BAC ) = \frac{3}{5}, \tan( \angle BAC ) = \dfrac{4}{3}$
Solution: $\overline{AB}$ is the hypotenuse $\overline{BC}$ is opposite to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle BAC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{BC}}{\overline{AB}}= \frac{\overline{BC}}{10} $ $ \overline{BC}=10 \cdot \sin( \angle BAC ) = 10 \cdot \frac{4}{5} = 8$